最近公共祖先 · Lowest Common Ancestor

［抄题］：

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

“The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

      _______3______
     /              \
    ___5__          ___1__
   /      \        /      \
   6      _2       0       8
         /  \
         7   4

For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

［思维问题］：

不知道子节点怎么用dc。直接对给出的p,q节点进行操作即可。

［一句话思路］：

左右分开 谁不空返回谁

［输入量］：空： 正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：

［画图］：

［一刷］：

(left != null && right != null) 时，返回的是root节点的结果，不需要再做递归运算了。是一个“合”的过程。

［二刷］：

［三刷］：

［四刷］：

［五刷］：

［总结］：

［复杂度］：Time complexity: O(n) Space complexity: O(n)

［英文数据结构，为什么不用别的数据结构］：

只有dc算法，没有数据结构

［其他解法］：

自己写traverse函数：不好，会形成全局变量

［Follow Up］：

有parent指针的：用对齐的方法做

［LC给出的题目变变变］：

Lowest Common Ancestor of a Binary Search Tree 一模一样的，约束条件没用，直接套。

public class Solution {
    /*
     * @param root: The root of the binary search tree.
     * @param A: A TreeNode in a Binary.
     * @param B: A TreeNode in a Binary.
     * @return: Return the least common ancestor(LCA) of the two nodes.
     */
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode A, TreeNode B) {
        if (root == null || A == root || B == root) {//
            return root;
        }
        //divide
        TreeNode left = lowestCommonAncestor(root.left, A, B);
        TreeNode right = lowestCommonAncestor(root.right, A, B);

        //conquer
        if (left != null && right != null) {
            return root;//
        }
        else if (left != null) {
            return left;
        }
        else if (right != null) {
            return right;
        }
        else {
            return null;
        }
    }
}